<4-2>
#5
2
f ( X )  1 x 3
Show f (1)  f (1)
but doesn’t exist c  ( 1,1)
f ' c( )
s.t.
2
f '( x)   x
3
Since
0
1
5
doesn’t exist at x=0
 f '( x) is not differentiable on (-1,1)
Thus we can’t use Rolle ’s Theorem.
#12
f ( x)  x3  x 1 ,[0, 2]
(1)f is a polynomial so it is continuous on [0,2]
2
(2) f '( x)  3x 1 exists on (0,2)
So f is differentiable on (0, 2), by M.V.T.
f (b)  f (a )
f '(c) 
  (a, b) sit
ba
f '(c)  3c2 1
f (2)  f (0) 9  ( 1)

5
20
2
3c  1  5
2
c
3c  4
2
4
2 3

3
3 , but only
c2 
4
3
2
3 is in (0,2)
#17
5
3
Let f ( x)  4x  x  2x 1
f (0)  1  0, f (1)  b  0
 p  (1,0) s.t. f ( p)  0
Suppose solutions are more than one.
f ( p)  0
f ( g ) 0 p  g
Since (1) f is polynomial, it is continuous
4
2
(2) f '(x)  70x  3x  2 exist on 
By Rolle’s Theorem. There is a number  ( p, g )
4
2
s.t. f '(c)  0  f '(x)  20x  3x  2  0
∴ f has exactly one real root.
#21
(a) suppose that a cubic polynomial p(x) has roots a1  a2  a3  a4 , so
P(a1)  P(a2 )  P(a3 )  P(a4 ) . By Rolle’s Theorem there are numbers a1, a2 , a3
with a1  c1  a2 , a2  c2  a3 , a3  c3  a4 , and P '(c1 )  P '(c2 )  P '(c3 )  0 .
Thus the second-degree polynomial P '( x) has three distinct real roots, which is
impossible.
(b) We prove by induction that a polynomial of degree n has at most n real roots. This
is certainly true for n=1. Suppose that the result is true for all polynomials of
degree n. And let P ( x ) be a polynomial of degree n+1, suppose that P ( x ) has
more than n+1 real roots, say a1  a2  a3...  an1  an2 , then
P(a1 )  P(a2 )  ...  P(an2 )  0 . By Rolle’s Theorem, there are real numbers
c1...cn1 with a1  c1  c2 , a2  c2  a3...an1  cn1  cn2 , and
P '(a)  P '(c2 )  ...P '(cn1)  0 . Thus the nth degree polynomial P '( x) has at least
n+1 roots. This contradiction shows that P ( x ) has at most n+1 real roots.
#23
By the Mean Value Theorem, f (4)  f (1)  f '(c)(4 1) for some c  (1, 4) . But for
every c  (1, 4) we have f '(1)  2 . Putting f '(1)  2 into the above question and
substituting f (1)  10 , we get f (4)  f (1)  f '(c)(4 1)  10  3 f '(c)  10  3  2  16 .
So the smallest possible value of f (4) is 16.
#25
f is a fuction which f '( x)  2 and f (0)  1 , f (2)  4 .
f '( x) exists on (  ,2) f is continuous on  .
f '(1) 
By M.V.T.   (0, 2) s.t.
5
 f '(c)   2.5  2
 f '( x)  2
2
f (2)  f (0)
20
∴ There doesn’t exist such a fuction.
#31
f ( x) 
1
x
g ( x) 
1
x
1
if x>0
1
x
if x<0
For x>0
f ( x)  g ( x)  f '( x)  g '( x)
For x<0
f '( x)  ( x 1 ) '   x 2
g '( x)  (1  x1 ) '   x 2
f '( x)  g '( x)
No. since f '( x)  g '( x)
∴ f ( x)  g( x)  c1
x (  ,0 )  ( 0 
,  )
on (, 0)
and f (x)  g(x)  c2 on (0, )
Not. f ( x)  g ( x)  c
on 
Since f '( x)  g '( x) at x=0
#35
Let f (t )  g (t )  h(t )
g , h are the position functions of two runners.
Suppose they spent T secs on the race.
Since g and h are differentiable on (0,T) and continuous on [0,T]
f (T )  f (0) 0  0
f '(c) 

0
T 0
T
By M.V.T.   (0, T ) s.t.
f ' (c )
g ' c
( ) h 'c ( )
0
 g ' (c ) h ' c( )
Hence they have same speed at t  c .